Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
const int M = 1e5 + 5;
const int INF = 1e9;
const int MOD = 1e9 + 7;
const int LG = 20;
const int MAX_A = 10;
int n, m, q, p[N][LG], Lv[N];
vector<int> g[N], people[N][LG];
vector<int> merge(vector<int>& a, vector<int>& b) {
  vector<int> res;
  for (int cnt = 0, i = 0, j = 0; cnt < MAX_A && (i < a.size() || j < b.size());
       cnt++) {
    if (i == a.size())
      res.push_back(b[j++]);
    else if (j == b.size())
      res.push_back(a[i++]);
    else if (a[i] < b[j])
      res.push_back(a[i++]);
    else
      res.push_back(b[j++]);
  }
  return res;
}
void dfs(int v, int par, int level) {
  Lv[v] = level;
  p[v][0] = par;
  for (auto u : g[v]) {
    if (u == par) continue;
    dfs(u, v, level + 1);
  }
}
void precompute() {
  for (int j = 1; j < LG; j++) {
    for (int i = 1; i <= n; i++) {
      p[i][j] = p[p[i][j - 1]][j - 1];
      people[i][j] = merge(people[i][j - 1], people[p[i][j - 1]][j - 1]);
    }
  }
}
vector<int> LCA(int u, int v) {
  vector<int> res;
  if (Lv[u] < Lv[v]) swap(u, v);
  int diff = Lv[u] - Lv[v];
  for (int i = LG - 1; i >= 0; i--) {
    if ((1 << i) & diff) {
      res = merge(res, people[u][i]);
      u = p[u][i];
    }
  }
  if (u == v) {
    res = merge(res, people[u][0]);
    return res;
  }
  for (int i = LG - 1; i >= 0; i--) {
    if (p[u][i] != p[v][i]) {
      res = merge(res, people[u][i]);
      u = p[u][i];
      res = merge(res, people[v][i]);
      v = p[v][i];
    }
  }
  res = merge(res, people[u][1]);
  res = merge(res, people[v][0]);
  return res;
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m >> q;
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  for (int i = 1; i <= m; i++) {
    int city;
    cin >> city;
    people[city][0].push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    sort(people[i][0].begin(), people[i][0].end());
  }
  dfs(1, 0, 1);
  precompute();
  while (q--) {
    int u, v, a;
    cin >> u >> v >> a;
    vector<int> ans = LCA(u, v);
    a = min(a, (int)ans.size());
    cout << a << ' ';
    for (int i = 0; i < a; i++) cout << ans[i] << ' ';
    cout << '\n';
  }
  return 0;
}